mae 323: review lecture 4 · mae 323: review lecture 4 2011 alex grishin mae 323 review of lecture...

Lecture 4MAE 323: Review

2011 Alex Grishin 1MAE 323 Review of Lecture 4

•Highlights of Lecture 2 (9/15/2011):

The (Cauchy) Stress Tensor Defined

•The Cauchy stress tensor arises naturally when one considers that the forces on

an arbitrary plan cut through a continuum can always be resolved into a

component normal and tangent to that plane.

•If only an infinitesimal portion of this plane is considered, the force over this

portion can be constant – as can the resulting pressure. This allows us to calculate

the pressure equilibrium in along coordinate directions as below

x

y

∆L

n

nx

ny

F

Tx

Ty

T

S

T1

S1

T2S2

1

2





•Since S and T are always at right angles, we can consider them independently. So,

to keep things simple, consider first a force on the inclined plane acting only

normal to the plane…

x

y

∆L

n

nx

ny

F

Tx

Ty T

T1

S1

T2S2

1

2

0

( )limA A∆ →

∆=

∆

FT

•Again, the traction, T

is defined by:





•Equilibrium in the X and Y directions follow according to:

xF∑

2 1

2 1

cos sin 0

cos sin

x

x

T L T L S L

T T S

θ θ

θ θ

∆ − ∆ − ∆ =

= +

yF∑

1 2

1 2

sin cos 0

sin cos

y

y

T L T L S L

T T S

θ θ

θ θ

∆ − ∆ − ∆ =

= +





•And moment equilibrium follows according to…:

x

y

T1

S1

T2

S2

1

2

PM∑•Sum moments about point P

T

P

∆y

∆x

2 1

2 1

0S x y S y x

S S

− ∆ ∆ + ∆ ∆ =

=

•So, we see that the tangential

“pressures” on the two sides parallel to

the x and y axes are always equal

•We’ll use this fact to swap the S1 and S2

terms in the previous summation…





•The preceding leads to…:

2 2x x yT T n S n= +

cos

sin

x

y

n

n

θ

θ

=

=

1 1y y xT Tn S n= +

•Or, as a matrix equation*:

2 2

1 1

x x

y y

T nT S

T nS T

=

(1)

•So the traction equilibria may expressed as (after

swapping the S2 and S1 terms because they’re

equal):





•The matrix of normal and tangential pressures is known as the Cauchy or

infinitesimal stress tensor. It’s two-dimensional form is shown below. It gets

this name because it is only valid for small increments of stress and

corresponding strain

2 2

1 1

xx xy

yx yy

T S

S T

σ σ

σ σ

=

•The preceding derivation could have been done in three dimensions following

the same procedure, but on a plane cut through a 3D infinitesimal cube (a

tetrahedron)…

•As we saw in slide 5: xy yxσ σ= So, the matrix is also symmetric





•Applying Newton’s Third Law in the x-direction gives:

x

y

σxx|x+∆x

σxy|x+∆x

σyx|y+∆y

σyy|y+∆y

σxx|x

σxy|x

σyy|y

σyx|y

∆x

∆ybx

by

| | | | 0

x

xx x xx x x xy y xy y y x

F

y y x x b x yσ σ σ σ+∆ +∆= − ∆ + ∆ − ∆ + ∆ + ∆ ∆ =

∑

•Rearranging and dividing by ∆x∆y yields:

| || |0

xy y y xy yxx x x xx xx

bx y

σ σσ σ +∆+∆−−

+ + =∆ ∆

•Taking the limit as ∆x→0, ∆y→0:

0xyxx

xb

x y

σσ ∂∂+ + =

∂ ∂





•Similarly, repeating the previous three steps in the y-direction yields:

•And, once again, even though we won’t go thru the steps, we will simply point

out that the full three dimensional equations can be obtained in a similar

manner, considering a three-dimensional cube element instead of a square.

Without showing the derivation, the equations are:

0

0

0

xyxx xzx

yx yy yz

y

zyzx zzz

bx y z

bx y z

bx y z

σσ σ

σ σ σ

σσ σ

∂∂ ∂+ + + =

∂ ∂ ∂

∂ ∂ ∂+ + + =

∂ ∂ ∂

∂∂ ∂+ + + =

∂ ∂ ∂

0yy xy

yb

y x

σ σ∂ ∂+ + =

∂ ∂





•The stress equilibrium equation is very useful and general (it applies to

all elastic continua)

•It may be used, for example to derive the differential equation for a

bar/truss as follows:

Example

0yy xy

yb

y x

σ σ∂ ∂+ + =

∂ ∂

F

A

x

y•Assume a

response in the

x-direction ONLY

•Also, assume

no body forces

(b=0)

(2)





•And for a single direction (x), Generalized Hooke’s Law

reduces to:

Eσ ε= (3)

where: du

dxε = (4)

Plugging (3) and (4) back into (2) yields:

2

20

uE

x

∂=

∂

Integrating (5) once gives:

(5)

u FE c

x A

∂= =

∂(6)





•Equation (6) says that a bar/truss is a “constant strain” element. This is

why linear shape functions are typically used for this element (nothing is

to be gained by using higher order shape functions)

•The most general form of the full stress equilibrium equation is:

0

0

0

xyxx xzx

yx yy yz

y

zyzx zzz

bx y z

bx y z

bx y z

σσ σ

σ σ σ

σσ σ

∂∂ ∂+ + + =

∂ ∂ ∂

∂ ∂ ∂+ + + =

∂ ∂ ∂

∂∂ ∂+ + + =

∂ ∂ ∂

,0

ij j ibσ + =

…which can be expressed more compactly with indicial notation:




The Small Strain Tensor

•Just as the displacement field is the natural counterpart to the external forces

in a structure, the strain field is the natural counterpart to the stress field in a

structure. So, in two spatial dimensions, we can expect four strain

components (three independent ones due to the symmetry in slide 7)

•We will designate the strain tensor as εεεε.

xx xy

yx yy

ε ε

ε ε

=

ε

And, in three dimensions:

xx xy xz

yx yy yz

zx zy zz

ε ε ε

ε ε ε

ε ε ε

=

ε





•Although it’s a little difficult to convey graphically, an infinitesimal element

undergoing nonzero εxx,εyy, and εyz represents a small perturbation from an

equilibrium configuration (the dashed square), such that the actual lengths, ∆x

and ∆y don’t change. With this in mind, the strain components are defined as:

x

y

,, |x yu v

,, |x x yu v +∆

,, |x y yu v +∆

,, |x x y yu v +∆ +∆

θy

θx

, ,

0

| |lim

x x y x y

xxx

u u u

x xε +∆

∆ →

− ∂= =

∆ ∂

, ,

0

| |lim

x y y x y

yyy

v v v

y yε +∆

∆ →

− ∂= =

∆ ∂

, ,

0

| |lim tan

x x y x y

x xx

v vv

x xθ θ+∆

∆ →

− ∂= = ≈

∂ ∆





And finally:

x

y

,, |x yu v

,, |x x yu v +∆

,, |x y yu v +∆

,, |x x y yu v +∆ +∆

θy

θx

, ,

0

| |lim tan

x y y x y

y yy

u uu

y yθ θ+∆

∆ →

− ∂= = ≈

∂ ∆

•The engineering shear strain is given by:

xy

u v

y xγ

∂ ∂= +

∂ ∂

•The tensor shear strain is ½ the

engineering shear strain:

1

2xy

u v

y xε

∂ ∂ = + ∂ ∂




The Small Strain Tensor•Thus, it should be clear that

•Once again, without going thru the details, we’ll point out that in three

dimensions, there are three additional unique strain components (letting w

stand for displacement in the z-direction)

andxy yx xy yx

γ γ ε ε= =

, , , ,

0

| |lim

x y z z x y z

zzy

v v w

z zε +∆

∆ →

− ∂= =

∆ ∂

1 1

2 2yz yz

v w

z yε γ

∂ ∂ = + = ∂ ∂

1 1

2 2xz xz

u w

z xε γ

∂ ∂ = + =

∂ ∂

•Note that the engineering strain represents the TOTAL shear

angle, whereas the tensor strain represents an average




Generalized Hooke’s Law for Isotropic Solids

1 0 0 0

1 0 0 0

1 0 0 0

0 0 0 (1 2 ) / 2 0 0(1 )(1 2 )

0 0 0 0 (1 2 ) / 2 0

0 0 0 0 0 (1 2 ) / 2

xxxx

yyyy

zzzz

xyxy

yzyz

xzzz

E

εσ ν ν ν

εσ ν ν ν

εσ ν ν ν

γσ νν ν

γσ ν

γσ ν

− − −

= −+ −

−

−

i ij jCσ ε=

xx

yy

zz

xy

yz

xz

σ

σ

σ

σ

σ

σ

=

σ

xx

yy

zz

xy

yz

xz

ε

ε

ε

γ

γ

γ

=

ε

•Through the use of Voigt Notation*, we can

express the constitutive relation between stress

and strain as…:…where the

stress and strain

tensors are now

represented as

column vectors




Poisson’s Effect

1xx xx

yy xx

zz xx

E

E

E

ε σ

νε σ

νε σ

=

= −

= −

•Suppose you have a uniaxial stress field in the x-direction. In

that case, the stress-strain relations for an isotropic solid

reduce to:

yy zz

xx xx

ε εν

ε ε= − = −

…and you can retrieve the following relations for Poisson’s

Ratio. This shows that it is a measure of lateral to axial

strain:




Poisson’s Effect•In other words, Poisson’s ratio may be thought of as a ratio of transverse to axial

strain (or contraction) under a uniaxial stress as shown in the figure below. It is

worth repeating: This a material property. Different materials will exhibit

different ratios of contraction under a given tensile load (the values may even be

negative or zero, but they must never exceed 0.5, in which case the material is

perfectly incompressible) :

xxε

yyε

zzεxx

σ

x

y

z

xxσ




Plane Stress•When analyzing bounded continua in two spatial dimensions, one can make

one of two assumptions. The first one is that all stress components in the third

dimension (let’s call it z) are zero. In other words:

•We plug this assumption into (4) :

0zz xz yz

σ σ σ= = =

1 0 0 0

1 0 0 0

1 0 0 0 01

0 0 0 2(1 ) 0 0

0 0 0 0 2(1 ) 0 0

0 0 0 0 0 2(1 ) 0

xx xx

yy yy

zz

xy xy

yz

zz

E

ε ν ν σ

ε ν ν σ

ε ν ν

γ ν σ

γ ν

γ ν

− − − − − −

= +

+

+

to yield:

1

2(1 )

xx xx yy

yy xx yy

xy xy

E

ε σ νσ

ε νσ σ

γ ν σ

−

= − + +

( )zz xx yyE

νε σ σ= − +

and:

(7) (8)

These

components

are zero




Plane Stress•Note that separation of the z-component strain reaction (due purely to

Poisson’s Effect) in (10) is integral to the plane stress assumption. We omit it

from the constitutive relation and apply it as a bookkeeping exercise after

calculating the in-plane reactions

•Re-writing (10) as a matrix equation:

1 01

1 0

0 0 2(1 )

xx xx

yy yy

xy xy

E

ε ν σ

ε ν σ

γ ν σ

− = − +

2

1 0

1 01

0 0 (1 ) / 2

xx xx

yy yy

xy xy

Eσ ν ε

σ ν εν

σ ν γ

= − −

•And, after inversion (using Mathematica):

(10)

And remember our bookkeeping:

( )zz xx yyE

νε σ σ= − +

(9)




Plane Strain•The other 2D assumption we could make is that all the strain components in the

z-direction are zero:

0zz xz yz

ε ε ε= = =

•This is a different kind of assumption because it can be explicitly enforced (it may

be stated as a boundary condition on the LHS of the governing differential

equation), whereas the plane stress assumption arises from an observed

approximation due to the absence of loads, gradients, and boundary conditions in

the z-direction.

•Simply plug the above assumptions into (4):

1 0 0 0

1 0 0 0

1 0 0 0 0

0 0 0 (1 2 ) / 2 0 0(1 )(1 2 )

0 0 0 0 (1 2 ) / 2 0 0

0 0 0 0 0 (1 2 ) / 2 0

xx xx

yy yy

zz

xy xy

yz

zz

E

σ ν ν ν ε

σ ν ν ν ε

σ ν ν ν

σ ν γν ν

σ ν

σ ν

− − −

= −+ −

−

−




Plane Strain

1 01

1 0(1 )(1 2 )

0 0 (1 2 ) / 2

xx xx

yy yy

yx xy

σ ν ν ε

σ ν ν εν ν

σ ν γ

− = − + − −

•This yields:

•And:

( )(1 )(1 2 )

zz xx yy

Eνσ ε ε

ν ν= +

+ −

Where we again remove the out-of-plane reaction purely due to Poisson’s Effect

and calculate it separately.

•As we mentioned before, this condition can be explicitly enforced in a structure.

Specifically, if we have a structure whose z-dimension is very long compared to the

other two. If we then fix the z-displacements at the extreme z-locations (the

“ends” of the structure) and ensure no load gradients in the z-direction, then the

plane strain assumption should hold at locations far from the ends.

(11)




Plane Strain (Axisymmetric Version)

•The planar axisymmetric formulation is another planar stress/strain formulation,

but we do not include it as a separate category because it is actually equivalent to

the plane strain formulation in cylindrical coordinates (all properties of plane strain

hold equally true of axisymmetric elements). Switching to a cylindrical reference

frame does introduce some interesting differences, however.

…all that is required to convert the

plane strain formulation to an

axisymmetric one is replace the y-

coordinate with the cylindrical z,

the x-coordinate with r, and z

coordinate with θ . This produces

a 4 x 4 constitutive matrix and 4 x 1

stress/strain vector:

( )(1 )(1 2 )

rr zz

Eθθ

νσ ε ε

ν ν= +

+ −

θ

r

z




Plane Strain (Axisymmetric Version)

( )(1 )(1 2 )

rr zz

Eθθ

νσ ε ε

ν ν= +

+ −

θ

r

z

( )( )

1 0

1 0

1 01 1 2

0 0 0 1/ 2

r r

z z

rz rz

Eθ θ

σ εν ν ν

σ εν ν ν

σ εν ν νν ν

σ γν

− − =

−+ − −




But What Physical Situations Produce Plane Stress or

Plane Strain?

Planar Model Problem

Large thickness

One end free

Plane strain

Both ends fixed in z

Plane strain

Small thickness

One end free

Plane stress

Both ends fixed in z

Plane strain




Conclusions

•The examples in Lecture 4 seemed to illustrate that in the absence of a singularity

associated with the free end, the plane stress and plane strain solutions are identical in

plane – the only difference being that the plane stress solution produces a nonzero axial

strain accompanied by a zero axial stress while the plane strain solution has a nonzero

axial stress accompanied by a zero axial strain

•This is further evidence that the plane stress solution is an approximation associated

with a free end condition. If the thickness of the part is thin enough, this end condition

dominates (leading to the plane stress condition). With sufficient thickness, parallel

planar sections far enough from this condition will not be influenced by it at all due to

Saint -Venant’s Principle (leading to the plane strain condition).

•We conclude by noting that even in thick parts satisfying the planar stress/strain

requirements, a free end with a singularity will have a solution lying somewhere between

the plane stress and plane strain conditions

mae 323: review lecture 4 · mae 323: review lecture 4 2011 alex grishin mae 323 review of lecture...

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